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Choosing between linear and quadratic drag, and solving linear drag in two dimensions: terminal speed $mg/b$, time constant $m/b$, and the charge in a magnetic field.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to decide which drag law applies, solve horizontal and vertical motion with linear drag, and find terminal speeds, time constants and coasting distances.
From Physics C you know that a resistive force opposes the velocity, that a drag proportional to speed leads to an exponential approach to terminal speed, and how to solve a separable first-order differential equation. The last lesson wrote Newton's second law in whatever coordinates suit the problem. This lesson asks which drag law applies to a given object and solves the linear case completely, in two dimensions.
| Term | What it means |
|---|---|
| Drag | The force of a fluid opposing an object's velocity, $\vec{f} = -f(v)\hat{v}$. |
| Linear drag | $f = bv$, from the fluid's viscosity; dominant for small, slow objects. |
| Quadratic drag | $f = cv^2$, from pushing fluid aside; dominant for large, fast objects. |
| Terminal speed | The speed at which drag balances the weight, $v_{\text{ter}} = mg/b$ for linear drag. |
| Time constant | $\tau = m/b$, the time for the gap to terminal speed to shrink by a factor of $e$. |
| Stokes's law | $b = 3\pi\eta D$ for a sphere of diameter $D$ in a fluid of viscosity $\eta$. |
| Cyclotron frequency | $\omega = qB/m$, the angular frequency of a charge circling in a magnetic field. |
An object moving through air feels a force opposite to its velocity. For a sphere of diameter $D$ it is well described by two terms,
$$f(v) = bv + cv^2, \qquad b = \beta D, \qquad c = \gamma D^2,$$
with $\beta = 1.6 \times 10^{-4}$ N s/m² and $\gamma = 0.25$ N s²/m⁴ for air at sea level. The linear term comes from viscosity, the fluid's internal friction; the quadratic term from the momentum given to the air pushed out of the way. Their ratio is
$$\frac{f_{\text{quad}}}{f_{\text{lin}}} = \frac{\gamma}{\beta}Dv = (1.6 \times 10^3\ \text{s/m}^2)\,Dv.$$
A baseball at $40$ m/s has a ratio of several thousand, so only the quadratic term matters. A micrometer oil drop at a fraction of a millimeter per second has a ratio far below one, so only the linear drag matters. With linear drag the second law is linear in the velocity: $m\dot{\vec{v}} = m\vec{g} - b\vec{v}$. Its $x$ and $y$ components separate, and each one solves with an exponential of time constant $\tau = m/b$.
Another way: picture
Imagine dropping a speck of pollen. For the first fraction of a millisecond it falls freely; then the air's viscous drag, growing with its speed, catches up with its weight. After that it drifts down at a steady crawl, its terminal speed. A puck sliding on a sticky surface does the same thing sideways: it slows by the same fraction in every interval $\tau$, and so covers a finite distance in an infinite time.
Another way: steps
The linear term is Stokes's law, $f = 3\pi\eta Dv$ for a sphere moving slowly through a fluid of viscosity $\eta$. For air, $\eta = 1.8 \times 10^{-5}$ Pa s, and $3\pi\eta = 1.7 \times 10^{-4}$, which is the constant $\beta$ above. The quadratic term is a momentum argument: the object sweeps out a volume $Av$ of air each second, gives it a speed of order $v$, and so pushes it with a force of order $\rho Av^2$.
Which term wins is measured by the Reynolds number, $\text{Re} = \rho vD/\eta$, the ratio of inertial to viscous effects. The ratio $f_{\text{quad}}/f_{\text{lin}}$ is essentially the Reynolds number divided by about $50$. Below a Reynolds number of about one, flow around a sphere is smooth and the linear law is excellent; above about a thousand, a turbulent wake forms and the quadratic law takes over.
With only linear drag acting horizontally, $m\dot{v}_x = -bv_x$. The rate of change of $v_x$ is proportional to $v_x$ itself, so $v_x = v_{x0}e^{-t/\tau}$ with $\tau = m/b$. Integrating again gives $x = v_{x0}\tau(1 - e^{-t/\tau})$.
Two consequences are worth stating. The object never stops, since the exponential is never zero, but it approaches a finite limit $x_\infty = v_{x0}\tau$. And the speed depends on position in a remarkably simple way: $v_x = v_{x0} - x/\tau$, falling linearly with the distance covered. Half the launch speed is lost when half the coasting distance has been covered, and half the launch speed is reached at $t = \tau\ln 2$.
Take downward as positive. Then $m\dot{v} = mg - bv$. The speed stops changing when $bv = mg$, at the terminal speed $v_{\text{ter}} = mg/b$. Writing the equation as $\dot{v} = -(v - v_{\text{ter}})/\tau$ shows the gap to terminal speed decays exponentially, so from rest
$$v(t) = v_{\text{ter}}\left(1 - e^{-t/\tau}\right), \qquad y(t) = v_{\text{ter}}t - v_{\text{ter}}\tau\left(1 - e^{-t/\tau}\right).$$
Notice $v_{\text{ter}} = g\tau$. A body that reaches terminal speed quickly also has a low terminal speed. For a $1$ μm oil drop, $\tau$ is a few microseconds and $v_{\text{ter}}$ a few tens of micrometers per second; for a pollen grain, $\tau$ is a fraction of a millisecond. After a few time constants the body is, for every practical purpose, at terminal speed.
Launch a projectile with velocity $(v_{x0}, v_{y0})$, taking up as positive. The horizontal equation is as above, and the vertical one gives $v_y = -v_{\text{ter}} + (v_{y0} + v_{\text{ter}})e^{-t/\tau}$. Integrating both and eliminating $t$ gives the trajectory
$$y = \frac{v_{y0} + v_{\text{ter}}}{v_{x0}}x + v_{\text{ter}}\tau\ln\left(1 - \frac{x}{v_{x0}\tau}\right).$$
The logarithm makes the path lopsided: it rises almost like a vacuum parabola but falls steeply, toward a vertical asymptote at $x = v_{x0}\tau$. Expanding for weak drag, the range is $R \approx R_{\text{vac}}(1 - \tfrac{4}{3}v_{y0}/v_{\text{ter}})$: drag matters when the launch speed is a noticeable fraction of the terminal speed. This is the first place in the course where a small parameter gives an approximate answer that can be checked against an exact one.
A charge $q$ moving in a uniform field $\vec{B} = B\hat{z}$ feels $q\vec{v} \times \vec{B}$, another force that depends on velocity. The components are $m\dot{v}_x = qBv_y$ and $m\dot{v}_y = -qBv_x$, which are coupled. Combining them into the complex number $\eta = v_x + iv_y$ gives $\dot{\eta} = -i\omega\eta$ with $\omega = qB/m$, whose solution is $\eta = \eta_0e^{-i\omega t}$.
Where linear drag had a real exponent and decayed, the magnetic force has an imaginary exponent and rotates: the velocity turns at the cyclotron frequency $\omega = qB/m$ without changing its magnitude, and the charge moves on a circle of radius $v/\omega$. Ernest Lawrence's cyclotron at Berkeley used exactly this: the frequency does not depend on the speed, so one alternating voltage kicks the particles on every half turn.
Checking an answer. For $t \ll \tau$ the solution must reduce to the vacuum result, since drag has had no time to act: expand $e^{-t/\tau} \approx 1 - t/\tau + t^2/2\tau^2$ and $y$ becomes $\tfrac{1}{2}gt^2$. For $t \gg \tau$ the velocity must be $v_{\text{ter}}$ downward and zero sideways. Units: $b$ is N s/m, which is kg/s, so $m/b$ is seconds. And a linear-drag answer for a baseball or a skydiver is wrong from the start, whatever its algebra: check the regime first.
For a sphere of density $\rho_s$ in the linear regime, the mass grows as $D^3$ and the drag coefficient as $D$, so $\tau = m/b = \rho_sD^2/18\eta$ grows as the square of the size. The terminal speed $g\tau$ scales the same way. Halve a droplet's diameter and it falls four times more slowly and settles four times sooner into that slow fall.
That square law is why size sorting happens in nature and in industry. A river drops its sand before its silt and its silt before its clay, because each grain settles at a speed set by $D^2$. Water treatment plants rely on the same effect in their settling basins, and a centrifuge speeds it up by replacing $g$ with $\omega^2r$, thousands of times larger, so that particles too small to settle under gravity in a reasonable time settle in minutes. Every one of these designs starts from the two numbers of this lesson, $\tau$ and $v_{\text{ter}}$.
At the University of Chicago in 1909, Robert Millikan and Harvey Fletcher measured the charge of the electron by watching oil drops about a micrometer across fall between two metal plates. With the plates uncharged, a drop fell at its terminal speed, a few hundredths of a millimeter per second, and timing it across a scale in the microscope gave its size through Stokes's law: $v_{\text{ter}} = \rho gD^2/18\eta$.
Switching on the electric field then held or lifted the drop, and the field needed gave its charge. Every charge came out as a whole multiple of $1.6 \times 10^{-19}$ C. The method works only because the drops are deep in the linear regime, where drag is exactly proportional to speed and the time constant is microseconds, so every drop is always at terminal speed. Millikan received the Nobel Prize in 1923.
Fog droplets are typically $10$ to $20$ μm across. Stokes's law gives terminal speeds of $3$ to $12$ mm/s, so a droplet near the top of a $100$ m fog bank over San Francisco Bay would take two to nine hours to settle in perfectly still air. Real fog clears when the sun evaporates the droplets or the wind carries them away, long before they fall.
Wildfire smoke is worse. Its particles are around $1$ μm, and because the settling speed goes as $D^2$, they fall a hundred times more slowly than fog droplets: about $30$ micrometers per second, or a tenth of a meter an hour. That is why smoke from fires in the western states can hang over cities for days, and travel across the country to the East Coast, and why the Environmental Protection Agency tracks it as PM2.5, the particles under $2.5$ μm that stay aloft and reach deep into the lungs.
Graphs of falling speed often show the curve rising and then turning flat, which suggests the body reaches terminal speed at some moment and then stops accelerating. With linear drag that never happens: the gap to terminal speed shrinks by the same factor $e$ in every time constant, so after $5\tau$ it is under one percent but never zero. In practice this is a distinction without a difference, but it matters when you set up the solution: there is no time at which to switch from one formula to another.
A second error is to use linear drag because the algebra is easy. The ratio $1.6 \times 10^3\,Dv$ decides the law, and for anything a person can throw it says quadratic.
A baseball has $D = 0.07$ m and moves at $5$ m/s. Compute the ratio of the two drag terms.
$\dfrac{f_{\text{quad}}}{f_{\text{lin}}} = 1.6 \times 10^3 \times 0.07 \times 5 = 560$
The ratio formula for spheres in air.
Decide the baseball's regime.
$560 \gg 1 \quad\Rightarrow\quad \text{quadratic}$
Even at a gentle toss the quadratic term dominates.
An oil drop has $D = 1.5$ μm and falls at $5 \times 10^{-5}$ m/s. Compute its ratio.
$1.6 \times 10^3 \times 1.5 \times 10^{-6} \times 5 \times 10^{-5} = 1.2 \times 10^{-7}$
Same formula, tiny size and speed.
Decide the drop's regime.
$1.2 \times 10^{-7} \ll 1 \quad\Rightarrow\quad \text{linear}$
Millikan's drops were deep in the linear regime.
Find the drop's linear coefficient.
$b = \beta D = 1.6 \times 10^{-4} \times 1.5 \times 10^{-6} = 2.4 \times 10^{-10}\ \text{N s/m}$
$b = \beta D$ with $\beta$ for air.
A dust grain of mass $2.0 \times 10^{-9}$ kg has $b = 4.0 \times 10^{-6}$ N s/m. Write the terminal speed.
$v_{\text{ter}} = \dfrac{mg}{b} = \dfrac{2.0 \times 10^{-9} \times 9.8}{4.0 \times 10^{-6}}$
Drag balances weight.
Evaluate the terminal speed.
$v_{\text{ter}} = 4.9 \times 10^{-3}\ \text{m/s}$
About five millimeters per second.
Find the time constant.
$\tau = \dfrac{m}{b} = \dfrac{2.0 \times 10^{-9}}{4.0 \times 10^{-6}} = 5.0 \times 10^{-4}\ \text{s}$
Half a millisecond.
Check the relation between them.
$g\tau = 9.8 \times 5.0 \times 10^{-4} = 4.9 \times 10^{-3}\ \text{m/s}$
$v_{\text{ter}} = g\tau$, as it must.
Find the speed after $1.0$ ms, two time constants.
$v = 4.9 \times 10^{-3}(1 - e^{-2}) = 4.9 \times 10^{-3} \times 0.865 = 4.2 \times 10^{-3}\ \text{m/s}$
Already eighty-six percent of terminal.
A $0.30$ kg puck with $b = 0.10$ kg/s is launched at $2.0$ m/s. Find the time constant.
$\tau = \dfrac{m}{b} = \dfrac{0.30}{0.10} = 3.0\ \text{s}$
Mass over drag coefficient.
Write the velocity.
$v = 2.0\,e^{-t/3.0}\ \text{m/s}$
Exponential decay from the launch speed.
Integrate for the position.
$x = 2.0 \times 3.0\left(1 - e^{-t/3.0}\right) = 6.0\left(1 - e^{-t/3.0}\right)\ \text{m}$
Starting at $x = 0$.
Find the limiting distance.
$x_\infty = v_0\tau = 6.0\ \text{m}$
The exponential vanishes as $t \to \infty$.
Find when the speed has halved.
$t = \tau\ln 2 = 3.0 \times 0.693 = 2.1\ \text{s}$
Setting $e^{-t/\tau} = \tfrac{1}{2}$.
Find where the speed has halved.
$x = v_0\tau\left(1 - \tfrac{1}{2}\right) = 3.0\ \text{m}$
Half the distance, as $v = v_0 - x/\tau$ predicts.
Write the balance of forces.
$v_{\text{ter}} = \dfrac{mg}{b}$
Drag equals weight.
Substitute the values.
$v_{\text{ter}} = \dfrac{3.0 \times 10^{-9} \times 10}{6.0 \times 10^{-6}}$
SI units.
Evaluate the terminal speed.
A tiny sphere dropped from rest in still air feels linear drag and has terminal speed $5$ cm/s. What fraction of that speed has it reached after one time constant $\tau = m/b$?
Complete the worked solution: a puck of mass $27$ kg is launched at $4$ m/s across a surface where the only horizontal force is a linear drag $bv$ with $b = 3$ kg/s. Find the time constant in seconds, the distance it coasts in meters, and the drag force at launch in newtons.
Divide the mass by the drag coefficient.
$\tau = \dfrac{m}{b} =$ k
The time for the speed to fall by a factor of $e$.
Integrate the velocity to find the total distance.
$x_{\max} = \displaystyle\int_0^\infty v_0e^{-t/\tau}\,dt = v_0\tau =$ x
The puck never stops, but the distance it covers converges.
Multiply the coefficient by the launch speed.
$f_0 = bv_0 =$ f
The drag is largest at launch and falls with the speed.
Check the distance against the launch force.
$\text{initial deceleration} \times \tau^2 = \text{coasting distance}$
Both come from the same two numbers, so they must agree.
Match each quantity for a sphere of diameter $D$ and mass $m$ to its formula.
| $mg/b$ | $m/b$ | $\beta D$ | $\gamma D^2$ | |
|---|---|---|---|---|
| terminal speed | ||||
| time constant | ||||
| linear coefficient | ||||
| quadratic coefficient |
For a sphere in air, $f_{\text{quad}}/f_{\text{lin}} = 1.6\,Dv$ with $D$ in millimeters and $v$ in m/s. Fill in the ratio for a sphere of diameter $4$ mm at $9$ m/s, at ten times that speed, and at half that speed.
| $f_{\text{quad}}/f_{\text{lin}}$ | |
|---|---|
| at the stated speed | |
| at ten times the speed | |
| at half the speed |
A body dropped from rest feels linear drag, with terminal speed $9$ m/s and time constant $6$ s. Write its downward speed $v$ as a formula in the time $t$, in seconds.
Answer:
A pollen grain of mass $7 \times 10^{-10}$ kg falls through still air with linear drag coefficient $b = 2 \times 10^{-6}$ N s/m. Taking $g = 10$ m/s², what is its terminal speed, in mm/s?
Answer: mm/s terminal speed
Fog over San Francisco Bay is made of water droplets about $20$ μm across. By Stokes's law such a droplet falls at a terminal speed of $12.10$ mm/s in still air. If the air were perfectly still, how long would a droplet take to settle through a $100$ m fog bank, in minutes?
Answer: minutes to settle
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A body dropped from rest feels linear drag, with terminal speed $5$ m/s and time constant $3$ s. Write its downward speed $v$ as a formula in the time $t$, in seconds.
Answer:
You can solve motion with linear drag. Explain to someone why fog droplets drift down for hours while raindrops fall in seconds.
18. Your turn: a sphere with linear drag has $m = 3.0 \times 10^{-9}$ kg and $b = 6.0 \times 10^{-6}$ N s/m. Find its terminal speed with $g = 10$ m/s²., step 3
$v_{\text{ter}} = 5.0 \times 10^{-3}\ \text{m/s}$
Five millimeters per second.